Suppose we have a mass *m* moving horizontally with velocity *v* on a frictionless surface. Directly in its path is a mass *M*, backed by a spring at resting length with spring constant *k*.

Now, what will be the maximum distance by which the spring is compressed if the collision between the masses is:

I) perfectly inelastic?

II) perfectly elastic?

I) Inelastic collision means that the masses “stick” when they collide. We use conservation of momentum: the initial momentum is *mv*, and so for post-collision velocity *v*_{f},

Now, to find the maximum compression of the spring, the simplest method is to use energy conservation: for displacement *x* from resting length, the potential energy in the spring is . Immediately after the collision, we have only the kinetic energy of the joined masses, so the energy is

.

Maximum compression of the spring occurs when the masses have zero velocity; the kinetic energy is zero, and so the energy is entirely potential, giving us:

.

II) A perfectly elastic collision conserves energy as well as momentum. Let *v*_{f} be the post-collision velocity of mass *m*, and *v*_{F} that of mass *M*. Conservation of momentum gives:

and conservation of energy gives:

Solving the first equation for *v*_{f},

Substituting into the energy result:

.

As *v*_{F}≠0, we have

(which is double the velocity of the inelastic case).

Note that we do not need *v*_{f}, as now only mass *M* is compressing the spring.

We have energy

,

and so at maximum compression,

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Tags: Collision, Conservation of Energy, Conservation of Momentum, Elastic, Friday Physics, Inelastic, physics

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